You invest €12 000 at a nominal rate of 4.8% per year, compounded quarterly. Find the value FV of the investment after 2 years.
PV = €12 0004.8% / 4n = 2
FV = €
2.Time value of moneyHard
A loan of €25 000 is repaid with equal monthly payments over 3 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.
P = €25 0002.4% / 12N = 36
A = €
3.NPV & IRRMedium
An investment of €30 000 gives these cash flows at the end of years 1 to 4: €11 000, €9 000, €9 000, €12 000. The cost of capital is 5%. Find the net present value NPV.
A project costs €20 000 and returns €10 500, €12 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 10 500CF2 = 12 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 3.5% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 3.5%n = 9
P = €
6.Bond pricing & yieldHard
A bond with 10 years to maturity (face value €1 000, annual coupon 6%) sells for €1 162. The issuer can call it in 4 years at a call price of €1 060. Find the yield to call YTC in percent.
F = €1 000coupon = 6%P = €1 162call price = €1 060call in 4 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €10 000 000 and debt worth €5 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 21%. Find the WACC.
A company has 5 000 000 shares at €34.00 each and 20 000 bonds trading at €1 050 each. The pre-tax cost of debt is 5% and the tax rate is 28%. The share’s beta is 1.1, the risk-free rate 3.5% and the market return 9%. Find the WACC.
β = 1.1rf = 3.5%rm = 9%rd = 5%T = 28%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 14, 60, 33, 21, 52.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 5. Find the z-score z of a score of 56.
μ = 60σ = 5x = 56
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3, 4p(x): 0.35, 0.1, 0.35, 0.05, 0.15
Var(X) =
12.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 3 correct answers?
n = 9p = 0.5k = 3
P =
13.Normal distributionMedium
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 30.8 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 30.8 min
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 500σ = 40P = 97.5%
x = h
15.Confidence intervalsMedium
A sample of 15 items has mean 116 and standard deviation s = 18. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 15x̄ = 116s = 1890%t(14) = 1.761
lower limit =
16.Confidence intervalsHard
The population standard deviation is 35. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 3?
σ = 35E = 395%
n =
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 8 students, mean score 72, standard deviation 9. Group 2: 6 students, mean score 59.7, standard deviation 12. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 8x̄₁ = 72s₁ = 9n₂ = 6x̄₂ = 59.7s₂ = 12
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 12. Find the minimum sample size n.
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 225Σx² = 55Σy² = 10271Σxy = 712
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 26) is made. Find its residual e = y − ŷ.
n = 5(9, 26)
e =
3Solutions
1.Time value of money
Answer: FV = 13202 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Monthly rate and number of payments
i = 0.024/12 = 0.002
N = 12·3 = 36
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.9306
A = 25 000·0.002/(1 − 0.9306) = €720.44
Answer
A = €720.44
3.NPV & IRR
Answer: NPV = 6286 €
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
Present value of each cash flow
CF0 = −30 000
CF1/(1 + 0.05)1 = 11 000/1.05 = 10 476.19
CF2/(1 + 0.05)2 = 9 000/1.103 = 8 163.27
CF3/(1 + 0.05)3 = 9 000/1.158 = 7 774.54
CF4/(1 + 0.05)4 = 12 000/1.216 = 9 872.43
Add them up
NPV = €6 286.42
Answer
NPV = €6 286.42
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 8.037 %
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
Bracket the IRR
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
NPV(5%) = 884.35
NPV(10%) = −537.19
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 5 + 5·884.35/(884.35 − (−537.19)) = 8.111%
exact: NPV(8.037%) = 0 → IRR = 8.037%
Answer
IRR = 8.037%
5.Bond pricing & yield
Answer: P = 1268 €
Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.
Per period
C = €35.00, i = 0.0175, N = 18
Present value of the coupons
€35.00·(1 − 0.7318)/0.0175 = €536.44
Present value of the face value
€1 000·0.7318 = €731.78
Price
P = €1 268.22
Answer
P = €1 268.22
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTC = 3.067 %
Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.
Equation for the yield
1162 = 60·(1 − (1 + y)−4)/y + 1060/(1 + y)4
Solve numerically
y = 3%: €1 164.82
y = 4%: €1 123.89
YTC = 3.067%
Answer
YTC = 3.067%
The yield to maturity is 4.002%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.
7.CAPM & WACC
Answer: WACC = 7.852 %
Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).
Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = (56 − 60)/5 = −0.8
Answer
z = −0.8
11.Probability
Answer: Var(X) = 1.948
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (30.8 − 32)/2.5 = −0.48
Upper tail
P = 1 − Φ(−0.48) = 1 − 0.3156 = 0.6844
Answer
P = 0.6844
14.Normal distribution
Answer: x = 578.4 h
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z for the percentile
Φ(z) = 0.975 → z = 1.96
Back to x
x = 500 + 1.96·40 = 578.4 h
Answer
x = 578.4 h
15.Confidence intervals
Answer: lower limit = 107.8
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Critical value
df = n − 1 = 14, t = 1.761
Margin of error
E = 1.761·18/√15 = 8.184
Interval
116 ± 8.184 → [107.8, 124.2]
Answer
lower limit = 107.8
16.Confidence intervals
Answer: n = 523
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Solve for n
n = (1.96·35/3)² = 522.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 523
Answer
n = 523
17.Hypothesis tests & confidence intervals
Answer: t = 2.199
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Pooled variance
sp² = (7·9² + 5·12²)/12 = 107.3
sp = 10.36
Standard error
SE = 10.36·√(1/8 + 1/6) = 5.593
Test statistic
t = (72 − 59.7)/5.593 = 2.199
Answer
t = 2.199
With df = 12, the two-sided 5% critical value is 2.179; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: n = 98
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((1.645 + 1.645)·12/4)² = 97.42
Round up
n = 98
Answer
n = 98
19.Linear regression & correlation
Answer: r = 0.9683
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Centred sums
Sxy = 712 − 15·225/5 = 37
Sxx = 55 − 15²/5 = 10
Syy = 10271 − 225²/5 = 146
Correlation coefficient
r = 37/√(10·146) = 0.9683
Answer
r = 0.9683
20.Linear regression & correlation
Answer: e = −1.9
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 30/5 = 6
ȳ = 93/5 = 18.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 124
Regression line
b = 124/40 = 3.1
a = 18.6 − 3.1·6 = 0
ŷ = 0 + 3.1·x
Predicted value
ŷ = 0 + 3.1·9 = 27.9
Residual
e = 26 − 27.9 = −1.9
Answer
e = −1.9
The point lies below the line.
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